Functions of Any Angle

Summary:
Up till now we’ve been
working with the trig functions in terms of the sides of triangles.
But now, we’ll extend the definitions for
any angle—actually, for any number.
That’s a little bit more abstract, but it makes the trig
functions useful for all sorts of problems that don’t involve
triangles at all.

Not Just Triangles Any More

The trig functions are sometimes called circular functions
because they’re intimately associated with circles. You’ve
already seen that, with all six
functions in a complicated diagram, but let’s reduce it to the
essentials.

Take a right triangle, and place one of the two acute angles
at the center of a circle, with the adjacent leg along the x
axis. The hypotenuse then runs from the center of the circle to a
point on the circumference, so the hypotenuse is a radius of the
circle. Or, you could say that it runs from the origin (0, 0) to
the point (x, y).

This comes right from the original definitions of sine and
cosine of an angle:

sin A = opposite/hypotenuse = y/r

cos A = adjacent/hypotenuse = x/r

But here’s the thing. If you take away the triangle, nothing
else really needs to change.
Think about an angle as a rotation around the center of a circle, and
you can extend the trig functions
to work for any angle.

The angle is always measured as a rotation from the positive
x axis.
Positive angles are counterclockwise rotations, and negative
angles are clockwise rotations.
This puts the acute angles, 0° to 90°, in Quadrant I, and
the obtuse angles, 90° to 180°, in Quadrant II.

In the diagram, the general angle A is drawn in
standard position, meaning that the vertex of the angle is at
the origin, and one side lies along the positive x axis.
I could have drawn any angle, but I just happened to draw one in
Q II.

You’re probably used to thinking of angles in terms of
degrees, but really they make more sense in radian measure.
Why is that? You
probably remember that the circumference of a circle is
2πr, where r is the radius. The
arc length for any angle A is
s = Ar, or the angle is
A = s/r. If you sweep out a full
rotation, all the way around the circle, you sweep out an arc of the
whole circumference, 2πr, and the angle must be
2πr/r, or just 2π.

Fractions of a full rotation are fractions of 2π. For
example, the 12 o’clock position is a quarter of the way
around the circle from the positive x axis, so the angle is
2π/4, which is π/2. The 6’oclock position is 3/4 of
the way around, going counterclockwise, so the angle is
(3/4) × 2π = 3π/2. But it’s also a quarter
of the way around, going clockwise, which is the negative direction,
so it’s also (−1/4) × 2π =
−π/2. (We’ll get into that more in
Periodic Functions, below.)

Half a rotation is 2π/2 = π, and of course
it’s also 360°/2 = 180°. Since
π = 180°, π/180° = 1 and
180°/π = 1. This is why, if you need to
convert between degrees and radians, you multiply by
π/180° or 180°/π—you’re multiplying by
1, so you’re changing the form of the number but not its value.
Incidentally, this technique of
multiplying by a carefully chosen form
of the number 1 lets you convert almost any units, like meters per
second into miles per hour, for instance.

Not only do you not really need a triangle, you don’t
even really need an angle. Just
think of a point moving along the circle, in either direction.
The “angle” is just the distance the point has traveled,
divided by the radius: A = s/r.
Counterclockwise motion is positive; clockwise is negative.

But it can get even simpler. Set the radius to 1, and you have
a unit circle. The moving point starts at (1, 0) and
travels a distance A. The point has coordinates
(x, y), which depend only on the distance
(counterclockwise or clockwise) from the point (1, 0).

The circle’s radius is 1—not 1 meter or 1
foot or 1 mile, just plain old 1. Likewise, A is just a pure
number. So really you can say that the arguments of the trig functions
are just pure numbers. And A can be any number, positive
or negative, so the trig functions can take any real number as their
arguments.

(21)On the unit circle:

sin A = y

cos A = x

This doesn’t change anything you’ve already learned.
It just extends the sine and cosine—instead of functions of an
angle 0° to 180°, they are functions of any real number.

The other function definitions don’t change at all. From
equation 3, you still have

tan A = sin A / cos A

which means that

tan A = y/x

and the other three functions are still defined as
reciprocals (equation 5).

Once again, there’s nothing new here: we’ve just
extended the original definitions to a larger domain.
sin A and cos A still have a range of
[−1, +1], just as they always did.

Why Bother?

It turns out that all kinds of physical processes vary
in terms of sines and cosines as functions of time:
length of the day over the course of a year;
vibrations of a
spring, or of atoms, or of electrons in atoms; voltage and current in
an AC circuit; pressure of sound waves, electric and magnetic field
strength as a beam of light propagates. Nearly every
periodic process can be described in terms of sines and
cosines.

Even static processes show sines and cosines: when forces are
operating at an angle, for instance.
There’s also
Snell’s Law,
which governs how light rays bend when passing from water to air, or
between any two media:
n1 sin θ1 =
n2 sin θ2,
where the n’s are characteristics of each material
called an index of refraction, and the angles are measured
from a perpendicular to the interface.

We won’t get too far into all that in these pages, but
this is why you need a solid grounding in trig to study physics,
engineering, and other fields.

Reference Angles

In the olden days, before calculators, every trig book had a
table of function values for acute angles, and if you needed the
function value for any other angle you would have to work it out using
a reference angle. Now, of course, we use an app or a pocket
calculator to get the function values, but the concept of a reference
angle is still useful in simplifying expressions.

The reference angle is the acute angle between the
x axis and the terminal side of the original angle.

Take a look at angle A, which is in Quadrant II. The
terminal side crosses the unit circle at
(x, y), where x = cos A
(which is negative) and y = sin A (which is
positive).

A vertical line from that point to the x axis has length
y, and it strikes the x axis −x units to the left
of the origin, at coordinates (x, 0). Why do I say −x
units left of the origin? Because x itself is negative, but all
distances are positive, so the positive distance must be minus the
negative x coordinate.

The reference angle is 180° − A, or
π − A. Why? Because the two angles together
equal 180° (π).

The diagram shows that y is not only sin A
but also sin(180° − A). Therefore
sin(180° − A) or
sin(π − A) = sin A.

What about x? Well,
cos(180° − A) = −x,
using the right triangle in the diagram. But
cos A = x (which is negative). Therefore
cos(180° − A) or
cos(π − A) =
−cos A.

There’s nothing special about angle A;
sin(π − A) = sin A and
cos(180° − A) = −cos A
for all angles A or all numbers A. We can get even
more general:
the six function values for any angle equal the functionvalues for its reference angle, give or take a minus sign.

What’s this “give or take” business? That’s what the next section
is about.

Signs of Function Values

Therefore the
signs of sine and cosine are the same as the signs of y and x. But you
know which quadrants have positive or negative y and x, so you know
which angles (or numbers) have positive or negative sines and cosines.
And since the other functions are defined in terms of the
sine and cosine, you also know where they are positive or
negative.

Spend a few minutes thinking about it, and draw some sketches.
For instance, is cos 300° positive or negative? Answer:
300° is in Q IV, which is in the right-hand half of the
circle. Therefore x is positive, and the cosine must be positive as
well. The reference angle is 60° (draw it!), so
cos 300° equals cos 60° and not
−cos 60°.

You can check your thinking against the chart that follows.
Don’t memorize the chart! Its purpose is to
show you how to reason out the signs of the function values whenever
you need them, not to make you waste storage space in your brain.

What about other angles? Well, 420° = 360° + 60°,
and therefore 420° ends in the same position in the circle as
60°—it’s just going once around the circle and then
an additional 60°. So 420° is in Q I, just like 60°.

You can analyze negative angles the same way. Take −45°.
That occupies the same place on the circle as
360° − 45°, which is +315°. −45° is in Q IV.

Examples: Function Values

As you’ve seen, for any function you get the numeric value by
considering the reference angle and the positive or negative sign by
looking where the angle is.

Example: What’s cos 240°?

Solution: Draw the angle and see that the reference angle is
60°; remember that the reference angle always goes to the x axis,
even if the y axis is closer. cos 60° = ½,
and therefore cos 240° will be ½, give or take a
minus sign. The angle is in Q III, where x is negative, and
therefore cos 240° is negative.
cos 240° = −½.

Example: What’s tan(−225°)?

Solution: Draw the angle and find the reference angle of
45°. tan 45° = 1. But −225° is in
Q II, where x is negative and y is positive; therefore y/x is
negative.
tan(−225°) = −1.

Identities for Related Angles

The techniques we worked out above can be generalized into a set
of identities. For instance, if two angles are supplements then you
can write one as A and the other as 180° − A or
π − A. You know that
one will be in Q I and the other in Q II, and you also know that
one will be the reference angle of the other. Therefore you know at
once that the sines of the two angles will be equal, and the cosines
of the two will be numerically equal but have opposite signs.

This diagram may help:

Here you see a unit circle (r = 1) with four identical
triangles. Their angles A are at the origin, arranged so that they’re
mirror images of each other, and their hypotenuses form radii of the
unit circle. Look at the triangle in Quadrant I. Since its
hypotenuse is 1, its other two sides are cos A and
sin A.

The other three triangles are the same size as the first, so their
sides must be the same length as the sides of the first triangle. But
you can also look at the other three radii as belonging to angles
180° − A in Quadrant II, 180° + A in Quadrant III,
and −A or 360° − A in Quadrant IV. The thin arcs near the center of the circle
trace the rotations. All of them have a
reference angle equal to A. From the symmetry,
you can immediately see things like
sin(180° + A) = −sin A and
cos(−A) = cos A.

The relations are summarized below. Don’t memorize them! Just
draw a diagram whenever you need them—it’s easiest if you use
a hypotenuse of 1. Soon you’ll find that you can quickly visualize the
triangles in your mind and you won’t even need to draw a diagram.
The identities for tangent are easy to derive: just divide sine by
cosine as usual.

sin(180° − A) = sin Asin(π − A) = sin A

cos(180° − A) = −cos Acos(π − A) = −cos A

tan(180° − A) = −tan A
tan(π − A) = −tan A

(22)

sin(180° + A) = −sin Asin(π + A) = −sin A

cos(180° + A) = −cos Acos(π + A) = −cos A

tan(180° + A) = tan A
tan(π + A) = tan A

sin(−A) = −sin A

cos(−A) = cos A

tan(−A) = −tan A

The formulas for the other functions aren’t
needed very often, but when you do need them they
drop right out of the definitions in equation 3 and
equation 5, plus the formulas just above this paragraph.
Two examples:

cot(180° + A) =
1/tan(180° + A) = 1/tan A⇒
cot(180° + A) = cot A

csc(−A) =
1/sin(−A) = 1/(−sin A) =
−1/sin A⇒
csc(−A) = −csc A

One final comment: Drawing pictures is helpful to avoid
memorizing things, but you might not consider it a rigorous proof of
equation 22.
Later, in Sum and Difference Formulas,
you’ll see how to prove these identities with algebra,
independent of any picture.

Periodic Functions (the Basics)

Once you think of the trig functions as based on the
(x, y) coordinates of a point on a circle, you can see
that
adding 2π (360°) to an angle or subtracting 2π (360°) is just moving once around the circle.
But if you move all the way around a circle, in either direction, you
end up where you started. So, for example, sin 495° =
sin(360° + 135°). (If you want to, you can go
further by noticing that 495° or 135° is in
Quadrant II, and the reference angle is 45°, so
sin 495° = sin 45°.)

But if you can go around the circle once, you can go around
the circle any number of times. So
adding or subtracting any multiple of 2π (360°)
doesn’t change the coordinates
(x, y), and therefore
doesn’t change the value of the sine or cosine.

For this reason we say that sine and cosine are
periodic functions with a period of 360° or 2π. Their values
repeat over and over again. Of course secant and cosecant, being
reciprocals of cosine and sine, must have the same period.

Solution: −85π/12 =
−7π − π/12. Because multiples of 2π
don’t change anything, you want to break out an even
multiple of 2π, so subtract π and add 12π/12 to
get

−85π/12 = −8π + 11π/12
⇒cos(−85π/12) =
cos(11π/12)

(Yes, you could write it as
−6π − 13π/12, but I think that makes it
harder to visualize. I like to keep any minus sign to the
multiple-of-2π part, which gets thrown away.)

Nearly there now!
But instructions were to use the smallest
positive angle, and you can do better than 11π/12.
The reference angle is π/12—sketch the angles if you need
to!
11π/12 is in Quadrant II, where cosines
are negative, so cos 11π/12 =
−cos π/12, and therefore
cos(−85π/12) = −cos π/12.

What about tangent and cotangent? They
are periodic too, but their period is 180° or π:
they repeat twice as fast as the others. You can see this from
equation 22: tan(180° + A) =
tan A says that the function values repeat every 180° or
π radians.

Summarizing all of this:

(23)For any integer k:

sin(A+2πk) = sin A

cos(A+2πk) = cos A

tan(A+πk) = tan A

Practice Problems

To get the most benefit from these problems, work them
without first looking at the solutions. Refer back to the chapter
text if you need to refresh your memory.

Recommendation: Work them on paper —
it’s harder to fool yourself about whether you really
understand a problem completely.

1
In which quadrant does the angle −868° occur? What about
42 radians? What are the signs of their sines, cosines, and tangents?

2Rewrite using the smallest possible positive angle of the
same trig function:
(a) sin 700°
(b) tan 780°
(c) cos(−390.5) (That’s
radians, since there’s no degree mark, but be careful! The angle
is −390.5, not −390.5π.)

3Rewrite as a function of just A:
(a) cos(720° − A)
(b) sin(43π + A)

BTW: Periodic Functions (Advanced)

You may want to skip this section, especially the first
time you read the chapter. I find the equations of periodic motion
interesting, but I have to admit that if you need them you’ll
get them in your other classes. In that case, if you find something
hard to follow, you may want to come back here for another
approach.

Period and Frequency

Almost every repetitive process is governed by sines and cosines.
You’ll learn about this in physics or engineering classes, or
you can look at one of the many Web sites out there. I like
Waves
in Tom Henderson’s
Physics Classroom
site.

Sine and cosine have periods of 2π, but obviously the
great majority of physical processes have different periods.

How can we modify a sine or cosine function to have
any desired period? Well, suppose you want a process that repeats
every second. That needs to go faster than sin t, which only
repeats every 2π seconds (about 6.3 s). To speed things up,
multiply t by some factor greater than 1.
In this case, use f(t) =
sin(2πt) for a period of 1 second.

Take a look at the plot at right. The blue curve is
sin t, and the red curve is sin(2πt). The
dots are spaced at 1 second horizontally, with a 2π-second
“yardstick” added. You can see that sin t repeats its
cycle once in 2π seconds, where sin(2πt)
repeats once per second. (You can click the image to enlarge it.)

That was just one specific example, but here’s the
general rule:
the period of a sine function is 2π over the coefficient of the variable.
So the period of sin(2πt) is 2π divided by
the coefficient, which is also 2π: 2π/(2π) = 1,
and the period is 1 second. What would a function with a period of 5
seconds look like? Well, 2π/5 = 0.4π, so the
coefficient needs to be 0.4π: sin(0.4πt).
At right I’ve graphed sin t in blue again, and
sin(0.4πt) in purple. You see that
sin(0.4πt) varies a bit more rapidly than
sin t, with a period of 5 seconds (5 tick marks) versus about
6.3 seconds.

Period answers the question, how long does it take one
complete waveform to pass a given point? The flip side of that,
frequency, answers the question, how many waveforms pass a
given point in one unit of time?
If the period is 1/4 second, then 4 waveforms pass each second, and
the frequency is 4 Hz (Hertz), or in older US lingo 4 cycles per
second.
If the period is 5 seconds, then 1/5
of a waveform passes in 1 second, and the frequency is 1/5 =
0.2 Hz.

Frequency is 1/period. Therefore, in
sin at, where the period is
2π/a, the frequency is a/2π.

Amplitude

The sine function varies between −1 and +1, but of
course most physical processes vary between other numbers. We use the
word amplitude for half the difference between the low point of
a wave and its high point (trough and crest).
The amplitude of a sine function is the coefficient of the function
(not the variable).
Suppose you have a waveform measuring
4″ (10 cm) from trough to crest. The amplitude is
therefore 2″ (5 cm), and the function is
2 sin(something) in inches, or
5 sin(something) in centimeters. (For this example
I’m ignoring the period of the wave.)

At the right I’ve graphed sin t in blue again, and
3 sin t in green. You can see that the two waveforms have
the same period of 2π, but the green one varies three times as
much in value. The amplitude is a measure of the energy in a wave;
specifically,
the energy in a wave is proportional to the square of its amplitude,
and
the proportion depends on the material that’s carrying the wave,
as well as other factors.

Phase

Finally, a waveform can have phase, meaning that
it’s shifted left or right from the normal position, or
forward or backward in time. Take a look at the plot to the right.
The blue curve is our old friend sin t, and the black curve is
sin(t + π/3). The two are
out of phase by π/3 or 1/6 of a cycle. We can also say
that the phase of the black curve is −π/3, because the
black curve lags behind the blue curve by π/3 units.

The second view is the same curves, just displaced vertically
so that you can see each curve more easily. Notice the crests of the
two curves, their troughs, the ascending crossings of the x
axis, and the descending crossings. In each case, the black curve is
π/3 behind the blue curve.

You might have noticed that I gradually stopped talking about sine and
cosine curves, and focused only on the sine curve. Why is that? Phase
is the key.

Look at the plot. The blue is sin t, displaced downward
two units. The brown is sin(t + π/2), and
the yellow is none other than our neglected friend cos t
(displaced upward two units). The cosine curve is just the sine curve,
with a phase shift. Pretty cool, huh?

Is this just an accident? No, and it’s easy to prove.
From equation 22,

General Waveform Equation

Example:
My bicycle tire has a diameter of 622 mm (about
24.5 in), and my normal cruising speed is around 10 mph
(16.21 km/h)
Suppose there’s a spot on the outer rim of the tire. Find the
height of that spot above the ground, as a function of time, assuming
that the spot is touching the ground at time 0.

Solution: I find it easiest to build up these
functions piece by piece.

The diameter of the wheel is 622/1000 = 0.622 m. The
radius is half that, 0.311 m. The circumference of the wheel is
0.622π m ≈ 1.954 m (76.9 in).

The wheel revolves 4.503/1.954 ≈ 2.304
times per second, so its frequency is 2.304 Hz.
(The period is 1/2.304 = 0.434 s, by the way.) Frequency is
the coefficient of t over 2π, so the coefficient must
be 2π times the frequency.
2π × 2.304 ≈14.48.

The amplitude is the radius of the wheel, 0.311 m
(12.24 in). Our tentative function is

h(t) =
0.311 sin 14.48t in meters,
12.24 sin 14.48t in inches

That doesn’t take account of phase or vertical shift, but
it’s a start.

Vertical shift is easy. The center of the wheel
isn’t on the ground; it’s 0.311 m
(12.24 in) above the ground, so now our function is

The last thing to consider is phase.
sin 0 = 0, so with a phase of 0 the spot would have to
start at 0.311 m (12.24 in) above the ground—the same
height as the center of the wheel, and toward the rear of the bicycle.
But we’re told that the spot
starts at ground level: h(0) = 0, not 0.311 m.
This means that the spot is lagging by π/2, a quarter of a
revolution. So our function becomes

h(t) =
0.311 + 0.311 sin(14.48t − π/2)
⇒

h(t) =
0.311 + 0.311 sin(14.48t − 1.571)
in meters

h(t) =
12.24 + 12.24 sin(14.48t − 1.571)
in inches

Let’s check that against equation 24. Frequency
is 14.48/2π = 2.304 Hz, check. Phase is
−1.571, which is a lag of π/2, a quarter cycle,
check. Amplitude is 0.311 m (radius of the wheel),
check. And vertical shift is also the radius of the wheel,
0.331 m, check.

What’s New

12 Dec 2016: Complete rewrite with new pictures,
highlighting the idea of generalizing the functions from angles in a
triangle to travel around a unit circle, and then to pure
numbers. Much new material about periodic
functions, including a new section
on equations of periodic motion.

23 Nov 2016:
Added quadrant numbers to a couple of the pictures.
Updated the mathematical notation, particularly the use of italics and spaces, to conform to the standard. I used Jukka Korpela’s comprehensive Writing Mathematical Expressions (2014, Suomen E-painos Oy), ISBN 978-952-6613-25-3.